# How do I solve 8x^2+2x=12x? Answered

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actually..i got my mom to help me (she's a math teacher) and after you move 12x over and set it equal to zero you get 8x^2 - 10x = 0 then you pull out 2x and get 2x(4x - 5) = 0 then you set both parts equal to zero and you get x=0 and x=4/5. just to clarify for everyone trying to solve this for me. thanks anyways for guys!! i really appreciate it. :)

also in reply to on of MahvishnuMan's post..0 can be the answer for any variable if the equation is set equal to zero and if there are no numbers without variables.  and also as a "modern day student" myself, we DO do a fair amount of graphing. and as a fix to my last post..i meant "thanks anyways guys!!"

The only possible outcome is x=0.<br /> <br /> Normally, your first move trying to solve the equation is to move all the variables to one side of the equation and leave nothing but constants on the other. The problem with this equation is that there are no constants that you can use to define the x's. Therefore, the only possibility that makes sense is x=0; it's the only number you can plug into x that works.<br />

Yeah, I know.  I read the solution that was provided.  However, since zero also solves the equation, I was at least partly correct.  If I would've graphed it, I would've seen that the plot crosses the x axis at two points.

I never claimed algebra was my strong suit.  It's been awhile since I used this stuff.

8x2 + 2x - 12x = 0<br /> 8x2 -10x = 0<br /> x(8x - 10) = 0<br /> <br /> x = 0<br /> <br /> 8x-10 = 0<br /> x = 10/8 = 1.25<br /> <br /> <strong>x = 0, 1.25</strong><br /> <br /> Check the solutions:<br /> <br /> 8(0)2 + 2(0) = 12(0)<br /> <strong>0 = 0 </strong><br /> <br /> 8(1.25)2 + 2(1.25) = 12(1.25)<br /> 12.5 + 2.5 = 15<br /> <strong>15 = 15</strong><br /> <br /> For a graphical solution, graph the parabola 8x2 + 2x and the line 12x, their intersections are the solutions to the problem. <br /> <br /> The degree of the equation -- the largest power of x -- states the number of zeros the equation will have. If the largest power of x is 2, then there are two solutions to the equation.

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Rewrite the equation as : some polynomial in x = 0
Like this:  8x2 -10x +0 =0

Now factor it:
8x2 -10x +0 =0
8x(x -10/8) =0
x(x-5/4) =0

And you can see the answer, x=0 or x=5/4, from inspection.

Check your work using Octave or Matlab. Write the poly as a vector of its coefficients, then feet that to the roots() function.
octave.exe:1> roots([8,-10,0])
ans =

1.25000
0.00000